Slide generator/hookes law problem

[Rijad Hadzic]

Homework Statement

A stiff spring with a spring constant of 1200 N/m is connected to a bar on the slide generator as show in figure P32.40 Assume the bar has length l = 60 x10^-2 m and mass .75 kg, and it slides without friction. The bar connects to a U shaped wire to form a loop that has width w = 40 x 10^-2 m and total resistance 25 ohms and that sits in uniform magnetic field B = .35 T. The bar is initially pulled 5x10^-2 m to the left and released so that it begins to oscillate. What is the induced current in the loop as a function of time I(t)? (Ignore any effects due to the magnetic force on the oscillating bar)

Homework Equations

The Attempt at a Solution

ε = Blv, I = ε/R

so I(t) = ( (BL) / R ) * v(t)

I(t) = .0084 * v(t)

I'm having a hard time finding the function of v(t)

I know from hookes law -kx = ma

-kx /m = a

-80 = a

-80 = d^2 x /dt

But where do I go from here? Sorry I am lost. Wasn't even sure to put this in the math forum or here..

So I know the acceleration is equal to a constant number, how do I make that in terms of time? Any help would be great..

 

[RedDelicious]

Why do you think the acceleration is constant? The system is oscillating, which means x is changing, but k and m are both constants. Using Hooke's Law, what does that tell you about a?

 

[Rijad Hadzic]

Why do you think the acceleration is constant? The system is oscillating, which means x is changing, but k and m are both constants. Using Hooke's Law, what does that tell you about a?

I see. so from

-kx /m = a

on the left side, although -k and m are constants, I was wrong to just plug in 5x10^-2 m for x because although that's what it starts as, x is actually a function of time..

so my expression for a should look like: -1600x(t) = a(t)

and x HAS to be a sin or cos function because its oscillating correct?

But I'm still confused on how to proceed. I need v(t), which means taking the integral of -1600x(t)

I'm still having trouble constructing a function for x(t) though.. I know its a sin function but I'm a bit confused. would -1600 be the amplitude of it?

 

[RedDelicious]

I see. so from

-kx /m = a

on the left side, although -k and m are constants, I was wrong to just plug in 5x10^-2 m for x because although that's what it starts as, x is actually a function of time..

so my expression for a should look like: -1600x(t) = a(t)

and x HAS to be a sin or cos function because its oscillating correct?

But I'm still confused on how to proceed. I need v(t), which means taking the integral of -1600x(t)

I'm still having trouble constructing a function for x(t) though.. I know its a sin function but I'm a bit confused. would -1600 be the amplitude of it?

Much better. Yes it will be a function of sine or cosine, depending on the initial conditions. No -1600, will not be the amplitude. You can immediately rule that out because it doesn't have units of distance.

The general solution is $$x(t) = A\cos{\omega t} + B\sin{\omega t}$$ where the amplitude is given by $\sqrt{A^2+B^2}$ and $\omega = \sqrt{k/m}$

You can simplify that a bit by using the initial conditions. $x(0) = x_0$ and $x'(0)=v(0) = 0$

 

[Rijad Hadzic]

Thanks for the reply. So far I've reached:

The position function as time in simple harmonic motion =

x(t) = 5x10^(-2) cos (ωt + φ)

5x10^-2 has to be the amplitude since that's how much it is pulled back from rest position

I also got φ = -pi because cosinverse(inital position / max position)

now I am having trouble finding ω. I don't think from the given info in the problem that I can find it actually... am I correct here?

Either way, taking the derivative, I get velocity..

-5x10^-2 ω sin(ωt + φ)

-5x10^-2 ω sin(ωt + -pi)

but -1600x(t) = a(t) so I have to multiply my velocity function by that constant right?

= 80ωsin(ωt - pi) = v(t)

because I(t) = .0084 v(t)

I(t) = 80ωsin(ωt - pi) * .0084

is my answer correct now? Also can you please tell me if what I stated (can't find ω with given info) is true??

Thanks for the help.